How do you simplify #(tan^2 (theta) csc^2 (theta)-1) / tan^2(theta)#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer P dilip_k Jul 16, 2016 1 Explanation: #(tan^2 theta csc^2 (theta)-1) / tan^2theta# #=(cancel(sin^2theta)/cos^2theta* 1/cancel(sin^2 theta)-1) / tan^2theta# #=(1/cos^2theta-1) / tan^2theta# #=(sec^2theta-1) / tan^2theta# #=tan^2theta / tan^2theta=1# Answer link Related questions What does it mean to prove a trigonometric identity? How do you prove #\csc \theta \times \tan \theta = \sec \theta#? How do you prove #(1-\cos^2 x)(1+\cot^2 x) = 1#? How do you show that #2 \sin x \cos x = \sin 2x#? is true for #(5pi)/6#? How do you prove that #sec xcot x = csc x#? How do you prove that #cos 2x(1 + tan 2x) = 1#? How do you prove that #(2sinx)/[secx(cos4x-sin4x)]=tan2x#? How do you verify the identity: #-cotx =(sin3x+sinx)/(cos3x-cosx)#? How do you prove that #(tanx+cosx)/(1+sinx)=secx#? How do you prove the identity #(sinx - cosx)/(sinx + cosx) = (2sin^2x-1)/(1+2sinxcosx)#? See all questions in Proving Identities Impact of this question 6542 views around the world You can reuse this answer Creative Commons License